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Integration by substitution for sin/cos products

  1. Sep 5, 2014 #1
    Ok so I might be doing something silly but I just don't understand what is going on here. So the integral:

    i = ∫ sin x (cos x)^3 dx

    First I say u = cos x. So du = - sin x dx.

    So now I have i = ∫ - u^3 du. Which gives: i = -(1/4)u^4 or -(1/4)(cos x)^4. Easy.

    But if I say u = sin x instead, this is what happens:

    So du = cos x dx. And I say i = ∫ sin x (cos x)^2 cos x dx.

    So I have i = ∫ u(1 - u^2) du or i = ∫ (u - u^3) du. WHAT.

    Why am I getting two different answers?? Which one is right and why?

    Thanks!
     
  2. jcsd
  3. Sep 5, 2014 #2

    HallsofIvy

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    Staff Emeritus
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    Actually, it doesn't. It gives -(1/4)(cos x)^4+ C for any constant C.

    You are not getting two different answers. The integral of "u- u^3" is (1/2)u^2- (1/4)u^4+ c which is equal to (1/2)(sin x)^2- (1/4)(sin x)^4+ c.

    See what happens if you replace (sin x)^2 by 1- (cos x)^2 in that last formula.
     
  4. Sep 5, 2014 #3
    Actually you get the same answer. In fact when you put [itex] u=\sin(x) [/itex] you get [tex]
    \int u-u^3\,du=\frac{u^2}{2}-\frac{u^4}{4}+c=\frac{\sin(x)^2}{2}-\frac{\sin(x)^4}{4}+c
    [/tex]
    now, if you work on the first result you got, you'll see that: [tex]
    -\frac{\cos(x)^4}{4}+c=-\frac{(\cos(x)^2)^2}{4}+c=-\frac{(1-\sin(x)^2)^2}{4}+c=\frac{\sin(x)^2}{2}-\frac{\sin(x)^4}{4}+\frac{1}{4}+c
    [/tex]
    The two results are the same because it is an indefinite integral and they differ by an arbitrary constant. It could also have been solved without substitution noticing that [itex] \cos(x)^2=1-\sin(x)^2 [/itex] and that [itex] D[f(g(x)]=f'(g(x)) g'(x) [/itex]


    Didn't realise we replied at the same time, sorry
     
  5. Sep 6, 2014 #4
    Oops! I forgot to put the constant of integration. And yeah I see now.. it was pretty silly haha. Thanks! :)
     
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